How can one prove that if x is an integer greater than 2, then x/(x-1) is a not an integer?
How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?
Intuitively I can see this is true but how to prove it?
elementary-number-theory
asked Nov 27 at 8:31
pirsquare
379319
add a comment |
How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?
Intuitively I can see this is true but how to prove it?
elementary-number-theory
asked Nov 27 at 8:31
pirsquare
379319
2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
add a comment |
How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?
Intuitively I can see this is true but how to prove it?
elementary-number-theory
asked Nov 27 at 8:31
pirsquare
379319
How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?
Intuitively I can see this is true but how to prove it?
elementary-number-theory
elementary-number-theory
asked Nov 27 at 8:31
pirsquare
379319
asked Nov 27 at 8:31
pirsquare
379319
edited Nov 27 at 8:40
asked Nov 27 at 8:31
pirsquare
379319
asked Nov 27 at 8:31
pirsquare
379319
asked Nov 27 at 8:31
pirsquare
379319
379319
2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
add a comment |
2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
2
2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39
add a comment |
3 Answers
3
active
oldest
votes
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
223k15156274
add a comment |
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.4k981179
add a comment |
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
19.2k2736
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015494%2fhow-can-one-prove-that-if-x-is-an-integer-greater-than-2-then-x-x-1-is-a-not%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
223k15156274
add a comment |
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
223k15156274
add a comment |
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
223k15156274
$$dfrac x{x-1}=1+dfrac1{x-1} $$
So, $x-1(ne0)$ must divide $1implies x-1=pm1$
answered Nov 27 at 8:34
lab bhattacharjee
223k15156274
answered Nov 27 at 8:34
lab bhattacharjee
223k15156274
answered Nov 27 at 8:34
lab bhattacharjee
223k15156274
answered Nov 27 at 8:34
lab bhattacharjee
223k15156274
223k15156274
add a comment |
add a comment |
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.4k981179
add a comment |
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.4k981179
add a comment |
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.4k981179
Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)
$$frac x{x-1}-1=frac 1{x-1}gt0$$
$$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$
So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.
answered Nov 27 at 8:47
Mark Bennet
80.4k981179
answered Nov 27 at 8:47
Mark Bennet
80.4k981179
answered Nov 27 at 8:47
Mark Bennet
80.4k981179
answered Nov 27 at 8:47
Mark Bennet
80.4k981179
80.4k981179
add a comment |
add a comment |
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
19.2k2736
add a comment |
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
19.2k2736
add a comment |
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
19.2k2736
The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.
Indeed, for $2<x_1<x_2$:
$$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$
answered Nov 27 at 9:06
farruhota
19.2k2736
answered Nov 27 at 9:06
farruhota
19.2k2736
answered Nov 27 at 9:06
farruhota
19.2k2736
answered Nov 27 at 9:06
farruhota
19.2k2736
19.2k2736
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015494%2fhow-can-one-prove-that-if-x-is-an-integer-greater-than-2-then-x-x-1-is-a-not%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33
oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39